To determine the genotypes of the parents with respect to red-green color blindness, we need to understand that red-green color blindness (which we can denote as "n") is an X-linked recessive trait.
Since color blindness is recessive, males (who have one X chromosome) will express the trait if their X chromosome carries the recessive allele (Xn). Females (who have two X chromosomes) will express the trait if both of their X chromosomes carry the recessive allele (XnXn). If a female has one normal vision allele (XN) and one colorblind allele (Xn), she will have normal vision but will be a carrier.
In this situation, two parents with normal color vision have a son who is colorblind.
- The son is colorblind (XnY), which means he received the Xn allele from his mother.
- The father (who is XY) cannot pass on an X chromosome to his son, so he must have the normal vision Y chromosome.
- Since the mother has to have at least one Xn allele to pass it to her son, she must be a carrier (XNXn).
The possible genotypes for the parents are therefore:
- Mother: XNXn (carrier for color blindness)
- Father: XNY (normal vision)
Thus, the correct answer is XNXn and XNY.